Class 10 Mathematics

Chapter 7 Solutions — Coordinate Geometry

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Solutions

Overview

Step-by-step NCERT solutions for Coordinate Geometry (Chapter 7, CBSE Class 10 Mathematics) — every question and answer worked out in full, not just the final result. You can also read the Coordinate Geometry textbook chapter.

Solved

What these solutions cover

All 20 questions in Coordinate Geometry are solved in the PDF. Here's what's inside, exercise by exercise:

Distance Formula

1.
Find the distance between the following pairs of points:
- (i) (2, 3) and (4, 1)
- (ii) (−5, 7) and (−1, 3)
- (iii) (a, b) and (−a, −b)
Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2?
Determine if the points (1, 5), (2, 3) and (−2, −11) are collinear.
Check whether (5, −2), (6, 4) and (7, −2) are the vertices of an isosceles triangle.
In a classroom, 4 friends are seated at the points A, B, C and D as shown in Fig. 7.8. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, "Don't you think ABCD is a square?" Chameli disagrees. Using distance formula, find which of them is correct. [From Fig. 7.8: A(3, 4), B(6, 7), C(9, 4), D(6, 1)]
6.
Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
- (i) (−1, −2), (1, 0), (−1, 2), (−3, 0)
- (ii) (−3, 5), (3, 1), (0, 3), (−1, −4)
- (iii) (4, 5), (7, 6), (4, 3), (1, 2)
Find the point on the x-axis which is equidistant from (2, −5) and (−2, 9).
Find the values of y for which the distance between the points P(2, −3) and Q(10, y) is 10 units.
If Q(0, 1) is equidistant from P(5, −3) and R(x, 6), find the values of x. Also find the distances QR and PR.
Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (−3, 4).

Section Formula

Find the coordinates of the point which divides the join of (−1, 7) and (4, −3) in the ratio 2 : 3.
Find the coordinates of the points of trisection of the line segment joining (4, −1) and (−2, −3).
To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1 m from each other along AD. Niharika runs 1/4 th the distance AD on the 2nd line and posts a green flag. Preet runs 1/5 th the distance AD on the eighth line and posts a red flag. What is the…
Find the ratio in which the line segment joining the points (−3, 10) and (6, −8) is divided by (−1, 6).
Find the ratio in which the line segment joining A(1, −5) and B(−4, 5) is divided by the x-axis. Also find the coordinates of the point of division.
If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.
Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, −3) and B is (1, 4).
If A and B are (−2, −2) and (2, −4), respectively, find the coordinates of P such that AP = (3/7)AB and P lies on the line segment AB.
Find the coordinates of the points which divide the line segment joining A(−2, 2) and B(2, 8) into four equal parts.
Find the area of a rhombus if its vertices are (3, 0), (4, 5), (−1, 4) and (−2, −1) taken in order. [Hint: Area of a rhombus = (1/2) × (product of its diagonals)]

Sample

Worked examples

3 questions from Coordinate Geometry, solved step by step.

Distance Formula

Find the distance between the following pairs of points: (i) (2, 3) and (4, 1) (ii) (−5, 7) and (−1, 3) (iii) (a, b) and (−a, −b)

Distance formula: d = √[(x₂ − x₁)² + (y₂ − y₁)²]

(i) Points: (2, 3) and (4, 1)

d = √[(4 − 2)² + (1 − 3)²]

= √[4 + 4] = √8 = 2√2

(ii) Points: (−5, 7) and (−1, 3)

d = √[(−1 − (−5))² + (3 − 7)²]

= √[(4)² + (−4)²]

= √[16 + 16] = √32 = 4√2

(iii) Points: (a, b) and (−a, −b)

d = √[(−a − a)² + (−b − b)²]

= √[(−2a)² + (−2b)²]

= √[4a² + 4b²] = 2√(a² + b²)

Answer: (i) 2√2 (ii) 4√2 (iii) 2√(a² + b²)

Section Formula

Find the coordinates of the point which divides the join of (−1, 7) and (4, −3) in the ratio 2 : 3.

Section formula: P dividing (x₁, y₁) and (x₂, y₂) in ratio m : n gives

x = (m·x₂ + n·x₁)/(m+n), y = (m·y₂ + n·y₁)/(m+n)

(x₁, y₁) = (−1, 7), (x₂, y₂) = (4, −3), m : n = 2 : 3

x = (2×4 + 3×(−1))/(2+3) = (8−3)/5 = 5/5 = 1

y = (2×(−3) + 3×7)/(2+3) = (−6+21)/5 = 15/5 = 3

∴ The required point is (1, 3).

Answer: (1, 3)

Distance Formula

Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2?

d = √[(36 − 0)² + (15 − 0)²]

= √[1296 + 225]

= √1521

Check: 39² = 1521, so √1521 = 39.

Yes, the distance between towns A and B = 39 km (with coordinates in km).

Answer: 39 units; the towns A and B are 39 km apart

All 20 questions in Coordinate Geometry are solved step by step in the Solutions PDF above.

Board papers

Coordinate Geometry in CBSE board papers

Tested in 111 questions across 23 Class 10 board papers (2020–2026). Open a set to see exactly how this chapter is asked:

Keep solving

Chapter 7 Solutions — Coordinate Geometry

Overview

What these solutions cover

Distance Formula

Section Formula

Worked examples

Find the distance between the following pairs of points: (i) (2, 3) and (4, 1) (ii) (−5, 7) and (−1, 3) (iii) (a, b) and (−a, −b)

Find the coordinates of the point which divides the join of (−1, 7) and (4, −3) in the ratio 2 : 3.

Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2?

Coordinate Geometry in CBSE board papers

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